# Question b7513

Nov 20, 2017

$\text{56.74 g}$

#### Explanation:

Firstly, we need to determine whether the quantity of ${C}_{2} {H}_{4}$ we have or the quantity of ${O}_{2}$ we have is the limiting factor in how much product can be produced.

The reaction proceeds as;

${C}_{2} {H}_{4} + 3 {O}_{2} \to 2 C {O}_{2} + 2 {H}_{2} O$

This means that for every 1 mol of ethene that reacts we need 3 mols of oxygen.

We are told that we have 32.10g of ethene. This translates to 1.15mols.

$\text{no. of mols" = "mass"/"Relative Atomic Mass" = "32.10g" / ("28 gmol"^-1) = "1.15 mols}$

This means that to combust 32.10g of ethene, we would need 3.45mols of Oxygen (3 x 1.15 mols). BUT we only have 64.00g of oxygen, so we need to know how many mols of oxygen we have.

$\text{no. of mols" = "mass"/"Relative Atomic Mass" = "64.00" / ("32 gmol"^-1) = "2 mols}$

2mols is LESS than the 3.45mols of ${O}_{2}$ that we would need to completely combust 32.10g of ethene. Therefore we have to understand that not all the ethene will be reacted in this combustion - but we need to work out how many mols of ethene will react.

If we need 1 mol of ethene for 3 mols of oxygen, then 2 mols of oxygen needs 0.67mols of ethene $\left(\frac{2}{3}\right)$.

Now, if you look at the original reaction; 1mol of ethene reacts to produce 2 mols of $C {O}_{2}$. So in this case, 1.34mols of $C {O}_{2}$ is produced ($2 \cdot 0.67$).

We can now work out the mass of $C {O}_{2}$ produced;

$\text{mass" = "number of mols" xx "relative atomic mass" = 1.34 xx 44 = "58.52 g}$

We are however told that the reaction proceeds with a 97%# yield. Therefore;

$\text{58.52 g" xx 0.97 = "56.74 g}$